228 Higher Engineering Mathematics
- (− 2 +j)
1
(^4) ⎡
⎣
Modulus 1. 223 ,arguments
38. 36 ◦, 128. 36 ◦,
218. 36 ◦and 308. 36 ◦
⎤
⎦
- (− 6 −j 5 )
1
(^2) [
Modulus 2. 795 ,arguments
109. 90 ◦, 289. 90 ◦
]
- ( 4 −j 3 )
− 2
[^3
Modulus 0. 3420 ,arguments 24. 58 ◦,
144. 58 ◦and 264. 58 ◦
]
- For a transmission line, the characteristic
impedanceZ 0 and the propagation coefficient
γare given by:
Z 0 =
√(
R+jωL
G+jωC
)
and
γ=
√
[(R+jωL)(G+jωC)]
GivenR= 25 ,L= 5 × 10 −^3 H,
G= 80 × 10 −^6 siemens, C= 0. 04 × 10 −^6 F
and ω= 2000 πrad/s, determine, in polar
form,Z 0 andγ.
[
Z 0 = 390. 2 ∠− 10. 43 ◦,
γ= 0. 1029 ∠ 61. 92 ◦
]
21.4 The exponential form of a
complex number
Certain mathematical functions may be expressed as
power series (for example, by Maclaurin’s series—see
Chapter 8), three examples being:
(i) ex= 1 +x+
x^2
2!
+
x^3
3!
+
x^4
4!
+
x^5
5!
+··· (1)
(ii) sinx=x−
x^3
3!
+
x^5
5!
−
x^7
7!
+··· (2)
(iii) cosx= 1 −
x^2
2!
+
x^4
4!
−
x^6
6!
+··· (3)
Replacingxin equation (1) by the imaginary number
jθgives:
ejθ= 1 +jθ+
(jθ)^2
2!
+
(jθ)^3
3!
+
(jθ)^4
4!
+
(jθ)^5
5!
+···
= 1 +jθ+
j^2 θ^2
2!
+
j^3 θ^3
3!
+
j^4 θ^4
4!
+
j^5 θ^5
5!
+···
By definition, j=
√
(− 1 ), hence j^2 =−1, j^3 =−j,
j^4 =1,j^5 =j, and so on.
Thus ejθ= 1 +jθ−
θ^2
2!
−j
θ^3
3!
+
θ^4
4!
+j
θ^5
5!
−···
Grouping real and imaginary terms gives:
ejθ=
(
1 −
θ^2
2!
+
θ^4
4!
−···
)
+j
(
θ−
θ^3
3!
+
θ^5
5!
−···
)
However, from equations (2) and (3):
(
1 −
θ^2
2!
+
θ^4
4!
−···
)
=cosθ
and
(
θ−
θ^3
3!
+
θ^5
5!
−···
)
=sinθ
Thus ejθ=cosθ+jsinθ (4)
Writing−θforθin equation (4), gives:
ej(−θ)=cos(−θ)+jsin(−θ)
However, cos(−θ)=cosθand sin(−θ)=−sinθ
Thus e−jθ=cosθ−jsinθ (5)
The polar form of a complex number z is:
z=r(cosθ+jsinθ). But, from equation (4),
cosθ+jsinθ=ejθ.
Therefore z=rejθ
When a complex number is written in this way, it is said
to be expressed inexponential form.
There are therefore three ways of expressing a com-
plex number:
- z=(a+jb), called Cartesian or rectangu-
lar form, - z=r(cosθ+jsinθ)orr∠θ,calledpolarform,and
- z=rejθcalledexponential form.
The exponential form is obtained from the polar form.
For example, 4∠ 30 ◦becomes 4ej
π
(^6) in exponential
form. (Note that inrejθ,θmust be in radians.)